Let $P(x,y)$ the assertion of the following F.E. $P(x,0)$ gives $f(x)^2=f(x^2)$, now $P(x,y)-P(x,-y)$ for $y \ne 0$ gives $f(-y)=-f(y)$. Now let $P(x,y)-P(y,x)$ for $x \ne y$, we get $2f(x^2+y^2)=\alpha(xf(x)+yf(y))$ so if $y=0$ here then we can get $2f(x)^2=2f(x^2)=\alpha xf(x)$ for all $x \ne 0$. Now for $x \ne y$ and $x,y \ne 0$ we have $2f(x^2+y^2)=2f(x^2)+2f(y^2)$ which means $f(x+y)=f(x)+f(y)$ for all reals $x \ne y$ (using odd in all $x \ne 0$) with $x,y \ne 0$ But now this means $-f(x)=f(-x)=f(x)-f(2x)$ for all $x \ne 0$ so $f(2x)=2f(x)$ for all $x \ne 0$ and therefore we have $f(x+y)=f(x)+f(y)$ for all $x,y \ne 0$. And in addition set $x=-y$ here for $y \ne 0$ then you get $f(0)=0$. Which means that $f$ is additive and odd in $\mathbb R$. In addition since $f(x^2)=f(x)^2$ we get that $f(x) \ge 0$ for all $x \ge 0$ thus $f$ is bounded below in all non-negatives so by Cauchy we can in fact get that $f(x)=cx$ for some constant $c$. Now clearly $c=0$ works so suppose $c \ne 0$. Remember that $2f(x)^2=\alpha xf(x)$ so in fact this gives $2c=\alpha$, now replacing in the original F.E. we see that $cx^2+cy^2=c^2x^2+c^2y^2$ so that means $c=1$ and therefore we get that $f(x)=0$ for all $x \in \mathbb R$ or $f(x)=x$ for all $x \in \mathbb R$ both work thus we are done .

MathLuis wrote: we get that $f(x)=0$ for all $x \in \mathbb R$ or $f(x)=x$ for all $x \in \mathbb R$ both work thus we are done . More precisely : Always a solution $f(x)=0\quad\forall x$ And if $\alpha=2$ (and only in this case), a second solution $f(x)=x\quad\forall x$

Shouldn't it be $f(x)=0 (x \neq 0)$ and $f(0)=t\in\{0,1\}$ or $f(x) = x$ when $\alpha=2$

Actually it's addictive on $\mathbb{R}_{\neq 0}$

MathLuis wrote: Let $P(x,y)$ the assertion of the following F.E. $P(x,0)$ gives $f(x)^2=f(x^2)$, now $P(x,y)-P(x,-y)$ for $y \ne 0$ gives $f(-y)=-f(y)$. Now let $P(x,y)-P(y,x)$ for $x \ne y$, we get $2f(x^2+y^2)=\alpha(xf(x)+yf(y))$ so if $y=0$ here then we can get $2f(x)^2=2f(x^2)=\alpha xf(x)$ for all $x \ne 0$. Now for $x \ne y$ and $x,y \ne 0$ we have $2f(x^2+y^2)=2f(x^2)+2f(y^2)$ which means $f(x+y)=f(x)+f(y)$ for all reals $x \ne y$ (using odd in all $x \ne 0$) with $x,y \ne 0$ But now this means $-f(x)=f(-x)=f(x)-f(2x)$ for all $x \ne 0$ so $f(2x)=2f(x)$ for all $x \ne 0$ and therefore we have $f(x+y)=f(x)+f(y)$ for all $x,y \ne 0$. And in addition set $x=-y$ here for $y \ne 0$ then you get $f(0)=0$. Which means that $f$ is additive and odd in $\mathbb R$. In addition since $f(x^2)=f(x)^2$ we get that $f(x) \ge 0$ for all $x \ge 0$ thus $f$ is bounded below in all non-negatives so by Cauchy we can in fact get that $f(x)=cx$ for some constant $c$. Now clearly $c=0$ works so suppose $c \ne 0$. Remember that $2f(x)^2=\alpha xf(x)$ so in fact this gives $2c=\alpha$, now replacing in the original F.E. we see that $cx^2+cy^2=c^2x^2+c^2y^2$ so that means $c=1$ and therefore we get that $f(x)=0$ for all $x \in \mathbb R$ or $f(x)=x$ for all $x \in \mathbb R$ both work thus we are done . f(x+y)=f(x)+f(y) for all x,y>0 and x,y\ne 0 and f(X+y)=f(X)+f(y) for all x,y<0 and x,y\ne 0,not f(x+y)=f(x)+f(y) for all x,y \in R